首页 \ 问答 \ CSS：获取表单标签以显示在表单元素之上（jsfiddle）(CSS: Get form label to display on top of form element (jsfiddle))

CSS：获取表单标签以显示在表单元素之上（jsfiddle）(CSS: Get form label to display on top of form element (jsfiddle))

 我试图让表单标签显示在每个表单元素的顶部。 我有jsfiddle这样做。  
 问题是，如何在每行保留其标签的同时获得多个表单元素？  
 你可以在我的例子中看到我在同一行上有“city，state，zip”，但由于我的CSS中的display: block ，它将它们分成不同的行。  
label { 
    font-weight:bold;
    display: block;
}

<form>

    <label for="name">Name</label>
    <input type="text" name="name" /><br />

    <label for="email">Email</label>
    <input type="text" name="email" /><br />

    <label for="city">City</label>
    <input type="text" name="city" /> 

    <label for="state">State</label>
    <input type="text" name="state" /> 

    <label for="zip">Zip</label>
    <input type="text" name="zip" /><br />

    <label for="last">Last</label>
    <input type="text" name="last" />

</form> 

I am trying to get the form label to display on top of each form element. I have the jsfiddle where this is done. 
The problem is, how do I get more than one form element per line while keeping its label directly above it?  
You can see in my example how I have "city, state, zip" all on the same line, but because of display: block in my CSS, it is breaking them up onto separate lines. 
label { 
    font-weight:bold;
    display: block;
}

<form>

    <label for="name">Name</label>
    <input type="text" name="name" /><br />

    <label for="email">Email</label>
    <input type="text" name="email" /><br />

    <label for="city">City</label>
    <input type="text" name="city" /> 

    <label for="state">State</label>
    <input type="text" name="state" /> 

    <label for="zip">Zip</label>
    <input type="text" name="zip" /><br />

    <label for="last">Last</label>
    <input type="text" name="last" />

</form> 

原文：https://stackoverflow.com/questions/13330678

更新时间：2021-06-26 22:06

最满意答案

 我刚从你的代码中注意到这一行：  
command.ExecuteNonQuery();
 
 只被召唤一次。 当你这样做时会发生什么：  
var connection = new SQLiteConnection("DataSource=:memory:;Version=3;New=True;");

connection.Open();

string sql = "CREATE TABLE  recently_viewed(movieid INTEGER,picture TEXT)";

SQLiteCommand command = new SQLiteCommand(sql, connection);
command.ExecuteNonQuery();

using ( command = connection.CreateCommand())
{
    command.CommandText = "INSERT into recently_viewed(movieid ,picture) values(@movieid,@picture)";
    command.Prepare();
    command.Parameters.AddWithValue("@movieid", id);
    command.Parameters.AddWithValue("@picture", picture);
    command.ExecuteNonQuery();
}


string   count_table = " SELECT count(*) FROM recently_viewed";

SQLiteCommand   com3 = new SQLiteCommand(count_table, connection);
com3.ExecuteNonQuery();

int   ctable = Convert.ToInt32(com3.ExecuteScalar().ToString());

Response.Write(ctable);

connection.Close();

I just noticed from your code that this line: 
command.ExecuteNonQuery();
 
is called only once. What happens when you do: 
var connection = new SQLiteConnection("DataSource=:memory:;Version=3;New=True;");

connection.Open();

string sql = "CREATE TABLE  recently_viewed(movieid INTEGER,picture TEXT)";

SQLiteCommand command = new SQLiteCommand(sql, connection);
command.ExecuteNonQuery();

using ( command = connection.CreateCommand())
{
    command.CommandText = "INSERT into recently_viewed(movieid ,picture) values(@movieid,@picture)";
    command.Prepare();
    command.Parameters.AddWithValue("@movieid", id);
    command.Parameters.AddWithValue("@picture", picture);
    command.ExecuteNonQuery();
}


string   count_table = " SELECT count(*) FROM recently_viewed";

SQLiteCommand   com3 = new SQLiteCommand(count_table, connection);
com3.ExecuteNonQuery();

int   ctable = Convert.ToInt32(com3.ExecuteScalar().ToString());

Response.Write(ctable);

connection.Close();

CSS：获取表单标签以显示在表单元素之上（jsfiddle）(CSS: Get form label to display on top of form element (jsfiddle))

最满意答案

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